It turns out that, even in the case of a maximally symmetric spacetime, the implementation of the degauging procedure for the higher-spin Cotton tensor is extremely non-trivial. For example, in AdS\(_3\)  the expressions for \(\mathfrak {C}_{\alpha (n)}(h)\)  with \(3\le n \le 6\)  are given by [1]}
\(\mathfrak {C}_{\alpha (3)}(h)=&\frac{1}{2^2}\bigg (3\big ({\cal D}_{\alpha }{}^{\beta }\big )^2{h}_{\alpha \beta (2)}+\big ({\cal Q}-9\mathcal {S}^2\big ){h}_{\alpha (3)}\bigg )~,\\\mathfrak {C}_{\alpha (4)}(h)=&\frac{1}{2^3}\bigg (4\big ({\cal D}_{\alpha }{}^{\beta }\big )^3h_{\alpha \beta (3)}+4( {\cal Q}- 20 {\cal S}^2) {\cal D}_{\alpha }{}^{\beta }{h}_{\alpha (3)\beta }\bigg )~,\\\mathfrak {C}_{\alpha (5)}(h)=&\frac{1}{2^4}\bigg (5\big ({\cal D}_{\alpha }{}^{\beta }\big )^4{h}_{\alpha \beta (4)}+10\big ({\cal Q}-33\mathcal {S}^2\big )\big ({\cal D}_{\alpha }{}^{\beta }\big )^2{h}_{\alpha (3)\beta (2)}\\&\phantom{extra}+\big ({\cal Q}-57\mathcal {S}^2\big )\big (\mathcal {Q}-25\mathcal {S}^2\big ) {h}_{\alpha (5)}\bigg )~,\\\mathfrak {C}_{\alpha (6)}(h)=&\frac{1}{2^5}\bigg (6\big ({\cal D}_{\alpha }{}^{\beta }\big )^5{h}_{\alpha \beta (5)}+20\big ({\cal Q}-48\mathcal {S}^2\big )\big ({\cal D}_{\alpha }{}^{\beta }\big )^3{h}_{\alpha (3)\beta (3)}\\&\phantom{extra}+\big (6{\cal Q}^2-704\mathcal {S}^2+18432\mathcal {S}^4{\cal Q}\big ){\cal D}_{\alpha }{}^{\beta }h_{\alpha (5)\beta }\bigg )~,\) 
